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The Power(ful) Equations (Posted on 2007-12-22) Difficulty: 2 of 5
If x and y are positive integers, L: LCM of x and y, G: GCD of x and y, then solve the following equations for x and y values:
1) xy=LG, L>x.
2) x2+y2 = L2+G2, L>x>G.

  Submitted by Praneeth    
Rating: 4.0000 (2 votes)
Solution: (Hide)
1) x=2, y=4.
2) No solution.

Explanation:
1) Let the powers of prime divisors of x and L in prime factorization form be xi and Li respectively.
=> xi=(G/y)*Li,but(x/L=G/y)
=> xi=(x/L)*Li,x is not equal to L.
=> xi < Li
=> yi = Li. So, the given eqn reduces to
=> xy=yx and x not equal to y.
This equation has only (2,4) as its solution.

2) x2+y2=L2+G2
Add both sides 2xy.
x2+y2+2xy=L2+G2+2LG (Since LG=xy)
=> x+y=L+G
Let x = G*a, y =G*b where a,b are positive and relative primes.
=> a+b=ab+1 => (1-a)(1-b)=0
As x is not equal to either L or G, either of a,b can't have 1 as its solution. So, no solution to 2nd equation.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
answerK Sengupta2007-12-23 06:44:30
SolutionLG problem!Chesca Ciprian2007-12-22 16:46:26
SolutionSolution, Part ISteve Herman2007-12-22 15:18:13
SolutionSolution, Part 2Steve Herman2007-12-22 14:51:16
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